Could anyone check my working out the Fourier series for $$ f(t)=\bigl(\begin{smallmatrix}
1 ...... 0

Question

Could anyone check my working out the Fourier series for $$ f(t)=\bigl(\begin{smallmatrix}
1 ...... 0<t< \frac{T}{2}\ 
-1 .......\frac{T}{2}<t<T
\end{smallmatrix}\bigr) $$

$$ a_n= \int_{0}^{T}f(t)cos(nwt)dt=0 $$ Which is obvious if you look at the graph.
$$ a_0= \int_{0}^{T}f(t)dt=0 $$Also evident.

With the sine terms:
$$ b_n= \int_{0}^{T}f(t)sin(nwt)dt=\int_{0}^{T/2}f(t)sin(nwt)dt+\int_{T/2}^{T}f(t)sin(nwt)dt $$
$$ =\int_{0}^{T/2}sin(nwt)dt-\int_{T/2}^{T}sin(nwt)dt =[-\frac{1}{nw}cos(nw(T/2))]-[[-\frac{1}{nw}cos(nw(T/2))]-[-\frac{1}{nw}cos(nwT)]] $$
$$ w=\frac{2 \pi}{T}$$

anonymous · Answer

The bit cut off was $$[-\frac{1}{nw}\cos(nw(T/2))]-[[-\frac{1}{nw}\cos(nw(T/2))]-[-\frac{1}{nw}\cos(nwT)]]$$
$$=-[\frac{1}{nw}\cos(nw(T/2))]+[\frac{1}{nw}\cos(nw(T/2))]-[\frac{1}{nw}\cos(nwT)]]$$

anonymous · Answer

$$=-[\frac{1}{nw}\cos(n\frac{2 \pi T}{2T})]+[\frac{1}{nw}\cos(n\frac{2 \pi T}{2T}))]-[\frac{1}{nw}\cos(n\frac{2 \pi T}{T})]]$$$$=-[\frac{1}{nw}\cos(n\pi)]+[\frac{1}{nw}\cos(n\pi))]-[\frac{1}{nw}\cos(n2 \pi)]]$$

experimentx · Answer

table after (19) http://mathworld.wolfram.com/FourierSeries.html

experimentx · Answer

here is better example for square wave http://mathworld.wolfram.com/FourierSeriesSquareWave.html

anonymous · Answer

Damn! I left in an unnecessary bracket!

anonymous · Answer

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